Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x-2y &= 8 \\ -8x+9y &= 6\end{align*}$
Begin by moving the $x$ -term in the second equation to the right side of the equation. $9y = 8x+6$ Divide both sides by $9$ to isolate $y$ $y = {\dfrac{8}{9}x + \dfrac{2}{3}}$ Substitute this expression for $y$ in the first equation. $8x-2({\dfrac{8}{9}x + \dfrac{2}{3}}) = 8$ $8x - \dfrac{16}{9}x - \dfrac{4}{3} = 8$ Simplify by combining terms, then solve for $x$ $\dfrac{56}{9}x - \dfrac{4}{3} = 8$ $\dfrac{56}{9}x = \dfrac{28}{3}$ $x = \dfrac{3}{2}$ Substitute $\dfrac{3}{2}$ for $x$ back into the top equation. $8( \dfrac{3}{2})-2y = 8$ $12-2y = 8$ $-2y = -4$ $y = 2$ The solution is $\enspace x = \dfrac{3}{2}, \enspace y = 2$.